Balancing Tor

$L_n(A⊗_R-)(B)≃L_n(-⊗_RB)(A)=\operatorname{Tor}^R_n(A,B)$ for all $n$. Proof: Choose a projective resolution $P_•\xrightarrow{ε}A$ and a projective resolution $Q_•\xrightarrow{η}B$. We can view $A,B$ as complexes concentrated in degree 0. Look at the double complexes $P⊗_RQ,A⊗_RQ,P⊗_RB$ we get morphisms of bicomplexes \begin{tikzcd} & P\otimes_R Q \arrow[dl,swap,"\varepsilon\otimes\mathrm{id}"] \arrow[dr,"\mathrm{id}\otimes\eta"] & \\ A\otimes_R Q & & P\otimes_R B \end{tikzcd} Since $\operatorname{Tot}$ is functorial, these maps in turn induce chain maps $$ f: \operatorname{Tot}\left(P_{\bullet} \otimes Q_{\bullet}\right) \rightarrow \operatorname{Tot}\left(M \otimes Q_{\bullet}\right) \cong M \otimes Q_{\bullet}, $$ and $$ g: \operatorname{Tot}\left(P_{\bullet} \otimes Q_{\bullet}\right) \rightarrow \operatorname{Tot}\left(P_{\bullet} \otimes N\right) \cong P_{\bullet} \otimes N . $$ We claim that $f$ and $g$ are quasi-isomorphisms, which would give the isomorphism on homology $L_n(A⊗_R-)(B)≃L_n(-⊗_RB)(A)=\operatorname{Tor}^R_n(A,B)$. Proof. To show that $\varepsilon\otimes\mathrm{id}$ is quasi-isomorphism, we need to show that $\operatorname{cone}(ε⊗Q)=\operatorname{Tot}C$ is acyclic. Since each $− ⊗ Q_q$ is an exact functor (because the $Q_q$ are projective, hence flat), the rows of $C$ are exact. Since $C$ is an upper half plane complex, the chain complex $\operatorname{Tot}(C_{••})$ is acyclic by the Acyclic Assembly Lemma. Thus, $\operatorname{cone}(f)$ is an acyclic chain complex, so by Corollary 8.4, we have that $f$ is a quasi-isomorphism.